What is the Molar Enthalpy of Neutralization (ÄHn) in kJ/mol if 5.00 moles of HCl(aq) (acid) neutralized 5.00 moles of NaOH(aq) (base) and released 2033 J of heat?
To calculate the molar enthalpy of neutralization (ΔHn) in kJ/mol, we need to first calculate the amount of heat released in Joules (J) during the neutralization reaction. Then, we can convert this value to kilojoules (kJ) and divide it by the number of moles of the limiting reactant to find the molar enthalpy of neutralization.
First, let's calculate the amount of heat released in joules using the given information:
Heat released during the reaction = 2033 J
Next, we need to convert this value to kilojoules by dividing it by 1000:
Heat released in kilojoules (kJ) = 2033 J / 1000 = 2.033 kJ
Now, let's determine the limiting reactant in the reaction. The limiting reactant is the one that is completely consumed, limiting the amount of product that can be formed. In this case, both HCl and NaOH are in a 1:1 mole ratio, so they are both completely consumed.
Since 5.00 moles of HCl and 5.00 moles of NaOH were used, we can see that both reactants are completely consumed, and there is no excess of either reactant. Therefore, we can choose either one of them to calculate the molar enthalpy of neutralization.
Finally, to find the molar enthalpy of neutralization, we divide the heat released (in kJ) by the number of moles of the limiting reactant. Since both HCl and NaOH are in a 1:1 mole ratio, we can choose either one to calculate the molar enthalpy of neutralization. Let's choose HCl:
Molar enthalpy of neutralization (ΔHn) = Heat released in kilojoules (kJ) / Moles of limiting reactant
= 2.033 kJ / 5.00 moles
= 0.4066 kJ/mol (rounded to four decimal places)
Therefore, the molar enthalpy of neutralization for the reaction is 0.4066 kJ/mol.